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If $x=\sec h^{-1} \frac{1}{2}+\tan h^{-1} \frac{1}{2}$, then $\cos h x=$
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Verified Answer
The correct answer is:
$\frac{4 \sqrt{3}+3}{3}$
Given that,
$\begin{aligned} x & =\operatorname{sech}^{-1} \frac{1}{2}+\tanh ^{-1} \frac{1}{2} \\ & =\ln (2+\sqrt{4-1})+\frac{1}{2} \ln \left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right) \\ & =\ln (\sqrt{3}+2)+\frac{1}{2} \ln \left(\frac{3 / 2}{1 / 2}\right) \\ & =\ln (\sqrt{3}+2)+\frac{1}{2} \ln 3 \\ & =\ln (\sqrt{3}+2)+\ln \sqrt{3} \\ & =\ln (2+\sqrt{3}) \sqrt{3}=\ln (3+2 \sqrt{3}) \\ \text { Now, cosh } x & =\frac{2 \sqrt{3}+3+\frac{1}{2 \sqrt{3}+3}}{2} \\ & =\frac{2 \sqrt{3}+3+2 \sqrt{3}-3 / 3}{2} \\ & =\frac{6 \sqrt{3}+9+2 \sqrt{3}-3}{6} \\ & =\frac{8 \sqrt{3}+6}{6}=\frac{4 \sqrt{3}+3}{3}\end{aligned}$
$\begin{aligned} x & =\operatorname{sech}^{-1} \frac{1}{2}+\tanh ^{-1} \frac{1}{2} \\ & =\ln (2+\sqrt{4-1})+\frac{1}{2} \ln \left(\frac{1+\frac{1}{2}}{1-\frac{1}{2}}\right) \\ & =\ln (\sqrt{3}+2)+\frac{1}{2} \ln \left(\frac{3 / 2}{1 / 2}\right) \\ & =\ln (\sqrt{3}+2)+\frac{1}{2} \ln 3 \\ & =\ln (\sqrt{3}+2)+\ln \sqrt{3} \\ & =\ln (2+\sqrt{3}) \sqrt{3}=\ln (3+2 \sqrt{3}) \\ \text { Now, cosh } x & =\frac{2 \sqrt{3}+3+\frac{1}{2 \sqrt{3}+3}}{2} \\ & =\frac{2 \sqrt{3}+3+2 \sqrt{3}-3 / 3}{2} \\ & =\frac{6 \sqrt{3}+9+2 \sqrt{3}-3}{6} \\ & =\frac{8 \sqrt{3}+6}{6}=\frac{4 \sqrt{3}+3}{3}\end{aligned}$
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